Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#define int long long
using namespace std;
const int N = 1e5 + 5, M = 1005;
int n, a[M][M], prime[N], vis[N], tot;
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
void init(int n)
{
	for(int i = 2; i <= n; i ++)
	{
		if(!vis[i]) prime[++ tot] = i;
		for(int j = 1; j <= tot && i * prime[j] <= n; j ++)
		{
			vis[i * prime[j]] = 1;
			if(i % prime[j] == 0) break;
		}
	}
}
int gcd(int a, int b) {return !b ? a : gcd(b, a % b);}
int lcm(int a, int b)
{
	if(!a || !b) return a + b;
	return a / gcd(a, b) * b;
}
signed main()
{
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	init(10000); n = read();
	if(n == 2) {puts("4 7\n23 10\n"); return 0;}
	for(int i = 1; i <= n; i ++)
		for(int j = ((i + 1) & 1) + 1; j <= n; j += 2)
			a[i][j] = prime[(i + j) / 2] * prime[n + (i - j) / 2 + (n + 1) / 2];
	for(int i = 1; i <= n; i ++)
		for(int j = (i & 1) + 1; j <= n; j += 2)
			a[i][j] = lcm(lcm(a[i - 1][j], a[i][j - 1]), lcm(a[i + 1][j], a[i][j + 1])) + 1;
	for(int i = 1; i <= n; i ++, printf("\n"))
		for(int j = 1; j <= n; j ++, printf(" "))
			printf("%lld", a[i][j]);
	fclose(stdin);
	fclose(stdout);
	return 0;
}